Oxidation number of sulphur in Na2S2O3 and iron in Fe3O4 come out fractional, how can that be?

I made a checklist of oxidation number rules (of course I did) and applied it to a few compounds from NCERT. It all worked nicely until I hit two cases.

In $Na_2S_2O_3$ : $2(+1) + 2x + 3(-2) = 0$ , so $x = +2$ . In $Fe_3O_4$ : $3x + 4(-2) = 0$ , so $x = +8/3$ . Both are fractions or at least odd looking. But my teacher said the actual structure of thiosulphate has one sulphur at +6 and one at -2. So is +2 wrong? And is $+8/3$ for iron real, can an atom have a fractional oxidation number when electrons move in whole numbers?

Also I'm confused by $CrO_5$ . If I do the arithmetic with oxygen as -2, I get $x + 5(-2) = 0$ , so Cr = +10. But chromium can't be +10, its group number is 6! How do I handle peroxides in these questions? Same issue in $H_2SO_5$ .

Last, disproportionation. I understand the definition (same element both oxidised and reduced) but I'd like two or three standard examples so I can recognise them in an assertion-reason question.

Thanks, and sorry for the wall of text, anxiety is high before the unit test.

7 answers & replies

[+]Tushar PawarDropperSample· 1mo ago

The rule-based number is the average oxidation number over all atoms of that element. The real structure can have the same element in different environments.

Thiosulphate: central S is bonded to three O and one S, terminal S is bonded only to the central S. Real states: central S = +6, terminal S = -2, average = (+6 − 2)/2 = +2. Both statements are true, +2 is the average.

$Fe_3O_4$ is a mixed oxide, $FeO\cdot Fe_2O_3$ . So one Fe is +2 and two are +3: (2 + 3 + 3)/3 = +8/3.

[+][–]Aarav Kulkarni11thSample· 1mo agoAuthor
"average" is the word I was missing. adding both of these under "traps" on my checklist. thank you

[+]Aman YadavDropperSample· 1mo ago

CrO5. Peroxide linkage means oxygen is -1, not -2. The structure has one O as oxo (-2) and four O in two peroxo groups (-1 each). $x + (-2) + 4(-1) = 0$ , so $x = +6$ . Fine. H2SO5: three normal O (-2) and two peroxo O (-1): $2(+1) + x + 3(-2) + 2(-1) = 0$ , so $x = +6$ . Rule: max oxidation number can't exceed the group number. If you get more, look for a peroxide.

[+]Ishaan Bose11thSample· 1mo ago

so fractional oxidation number means the atom actually has a fractional charge? like fe in fe3o4 really lost 2.67 electrons each? chemistry is cursed

[+]Nisha Verma12thSample· 1mo ago

No, Ishaan. Oxidation number is bookkeeping, not real charge. Electrons move in whole numbers. A fraction just means the atoms of that element are in different environments and you've taken the average, like in Fe3O4 (two Fe³⁺, one Fe²⁺) or Na2S4O6 (average +2.5). If an NCERT question asks for the oxidation state in an actual structure, use the structure. If it asks for the "oxidation number" using rules, you'd give the average.

[+][–]Ishaan Bose11thSample· 1mo ago
ok bookkeeping not real charge, makes sense. still cursed though

[+]Riya Singh12thSample· 1mo ago

Disproportionation examples:

$2H_2O_2 \rightarrow 2H_2O + O_2$ (oxygen goes from -1 to -2 and to 0)
$Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$ (chlorine goes from 0 to -1 and to +1)
$P_4$ in alkali gives $PH_3$ and $H_2PO_2^-$

Recognise by one reactant, the same element appears in a higher and a lower state in the products. Also keep in mind: oxygen is +2 in $OF_2$ , -1 in peroxides and -1/2 in superoxides like $KO_2$ .

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